Problem: Find the limit as $x$ approaches negative infinity. $\lim_{x\to-\infty}\dfrac{\sqrt{x^8-5x^3}}{3x^4+4}=$
Let's find this limit directly. To do that, we will want to divide both the numerator and the denominator by the same quantity, in a way that will help us derive the limit. Since the leading term of the denominator is $x^4$, let's divide by $x^4$. In the numerator, let's divide by $\sqrt{x^8}$, since for any value, $x^4=\sqrt{x^8}$. $\begin{aligned} &\phantom{=}\lim_{x\to-\infty}\dfrac{\sqrt{x^8-5x^3}}{3x^4+4} \\\\ &=\lim_{x\to-\infty}\dfrac{\dfrac{\sqrt{x^8-5x^3}}{\sqrt{x^8}}}{\dfrac{3x^4+4}{x^4}} \gray{\text{Divide sides by }x^4=\sqrt{x^8}} \end{aligned}$ Now let's continue by simplifying the expression and using the fact that for any nonzero number $k$ and positive power $n$, the limit $\lim_{x\to-\infty}\dfrac{k}{x^n}$ is equal to $0$. $\begin{aligned} &=\lim_{x\to-\infty}\dfrac{\sqrt{\dfrac{1\cancel{x^8}}{\cancel{x^8}}-\dfrac{5\cancel {x^3}}{\cancel {x^3}\cdot x^5}}}{\dfrac{3\cancel{x^4}}{\cancel{x^4}}+\dfrac{4}{x^4}} \\\\ &=\lim_{x\to-\infty}\dfrac{\sqrt{1-\dfrac{5}{x^5}}}{3+\dfrac{4}{x^4}} \\\\ &=\lim_{x\to-\infty}\dfrac{\sqrt{1-0}}{3+0} \gray{\lim_{x\to-\infty}\dfrac{k}{x^n}=0} \\\\ &=\dfrac{\sqrt{1}}{3} \\\\ &=\dfrac{1}{3} \end{aligned}$ In conclusion, $\lim_{x\to-\infty}\dfrac{\sqrt{x^8-5x^3}}{3x^4+4}=\dfrac{1}{3}$.